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Math Help - Prime number sum

  1. #1
    Junior Member
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    Prime number sum

    From a proof I have reason to believe the following inequality holds

    \displaystyle \sum_{\substack{ p\leq x \\ p| q}} 1 \ll \log q

    where p signifies summing only over primes. But I cannot prove it.

    (The full argument used is: \displaystyle \sum_{\substack{ p\leq x \\ p| q}} \bigg[\frac{\log x}{\log p}\bigg]\log p \ll (\log x)\log q )
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  2. #2
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
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    If I understand correctly, you are trying to bound the number of distinct prime divisors for a number q by \log q? If this is correct, then you can find a loose bound on the number of distinct prime divisors.

    Take k such that 2^{k-1}\leq q< 2^k. Then, q can have at most k-1 distinct prime divisors. Taking the logarithm gives k-1\leq \log_2 q\leq k. So in conclusion,

    \displaystyle \sum_{p|q} 1\leq k-1\leq \log_2 q

    where p is prime. This bound is actually sharp: take q=2.
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